Sixteen golfers, seven rounds. What are optimum combinations for each golfer to play with as many different golfers as possible?

Many thanks,
Bob

Hi Bob

As in the case of your previous problem there is no really nice solution. To construct a solution for sixteen golfers and seven rounds you can modify my previous solution. I deleted week 8, since you only need seven weeks, and then added the four additional players, 13, 14, 15, and 16 as a foursome in the first week. In each successive week the fourth foursome includes a pair of the new golfers (there are exactly six such pairs) and a pair of the golfers from the original list. The pairs chosen from the original golfers are pairs that only appeared together twice in the first schedule. Thus in this new schedule every pair appears together either 2 or 3 times and it avoids any triple being repeated.